Solving 2 differential equations

oxxiissiixxo
Messages
27
Reaction score
0
dx/dt=ay and dy/dt=bx where x and y are function of t [x(t) and y(t)] and a and b are constant.

1) show what x and y satisfy the equation for a hyperbola: y^2-(b/a)*x^2=(y_0)^2-(b/a)*(x_0)^2

2) suppose at some time t_s, the point (x(t_s),y(t_s)) lies on the upper branch of hyperbola, show that: y(t_s)>sqrt(b/a)*x(t_s)

I dun know whether i am doing it right.

First, in integrate both equations,

dx/dt=ay >>> x/y+C_1=at+C_2 >>> x/y+C_5=at

dy/dt=bx >>> y/x+C_3=bt+C_4 >>> y/x+C_6=bt

then I say t = 0 and so

x/y+C_5=at >>> C_5=-x_0/y_0

y/x+C_6=bt >>> C_6=-y_0/x_0

then i say this happens only when C_5 and C_6 are 0

then going back to

x/y+C_5=at >>> x/y=at

y/x+C_6=bt >>> y/x=bt and isolating t to yield

y^2-(b/a)*x^2=0

and when t=0

y_0^2-(b/a)*x_0^2=0

so y^2-(b/a)*x^2=y_0^2-(b/a)*x_0^2

am i right about it?

and can somebody give me some hints to deal with the second problem? thank you.
 
Physics news on Phys.org
hi oxxiissiixxo! :smile:

(try using the X2 button just above the Reply box :wink:)
oxxiissiixxo said:
dx/dt=ay >>> x/y+C_1=at+C_2 >>> x/y+C_5=at

dy/dt=bx >>> y/x+C_3=bt+C_4 >>> y/x+C_6=bt

i've no idea what you're doing here :redface:

you can't possibly integrate those equations​

hint: try differentiating the equation they've given you :wink:
1) show what x and y satisfy the equation for a hyperbola: y^2-(b/a)*x^2=(y_0)^2-(b/a)*(x_0)^2
 
You can't integrate the equations like that. dx/y isn't d(x/y). I suggest you differentiate the hyperbola equation.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top